Problem: A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".

You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.

You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.

Determine if it is possible to obtain an RBS after these replacements.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.

Output

For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}. 

You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Example

Input


5
()
(?)
(??)
??()
)?(?


Output


YES
NO
YES
YES
NO

Note

In the first test case, the sequence is already an RBS.

In the third test case, you can obtain an RBS as follows: ()() or (()).

In the fourth test case, you can obtain an RBS as follows: ()().

Solution (in PYTHON3):

Solution:


def generate(s):
    if len(s)==1 or s[0]==')' or s[-1]=='(':
        return 'NO'
    
    counto=0
    countc=0
    countq=0
    # ptr=0
    for i in s:
        if i=='(':
            counto+=1
            
        if i==')':
            countc+=1
            
        if i=='?':
            countq+=1
        # ptr+=1
        # print(ptr)
        # print(counto,countc,countq)
        # print('--------------')
    if counto>countc:
        if counto==countc+countq:
            return 'YES'
        else:
            return 'NO'
    
    if counto<countc:
        if (counto+countq)==countc:
            return 'YES'
        else:
            return 'NO'
    
    if counto==countc and countq%2==0:
        return 'YES'
    else: 
        return 'NO'
    

t=int(input())
ans=[]
for _ in range(t):
    s=list(input())
    # print(s)
    ans.append(generate(s))

for i in range(t):
    print(ans[i])